Problem: $h(t)=-(t-7)^2+4$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
Explanation: $\begin{aligned} -(t-7)^2+4&=0 \\\\ (t-7)^2&=4 \\\\ \sqrt{(t-7)^2}&=\sqrt{4} \\\\ t-7&=\pm 2 \\\\ t&=\pm2+7 \\\\ t={5}&\text{ or }t={9} \end{aligned}$ $h(t)$ is given in vertex form: $h(t)=-(t-{7})^2+{4}$ So the vertex of the parabola is at $({7},{4})$. In conclusion, $\begin{aligned} \text{smaller }t&=5 \\\\ \text{larger }t&=9 \end{aligned}$ The vertex of the parabola is at $(7,4)$